AP Unit 9 Solutions

Conserve vector momentum: m(300,400) = m(300,0) + m(vBx,vBy), so vB = (0,400) m/s.

Initial kinetic energy is (m/2)(300²+400²). Final kinetic energy is (m/2)(300²+400²), the same.

A changes normal momentum from +m(300) to −m(300): Δpx = −2m(300) = −2.4 × 10⁻²³ kg·m/s. The wall gets +2.4 × 10⁻²³ N·s.

Average wall force = (2.4 × 10⁻²³)(5.0 × 10²²) = 1.2 N. Pressure = 1.2/0.040 = 30 Pa.

Imagine a small sensor or surface inside the gas. Atoms would transfer normal momentum to both sides.

Equal forces on opposing faces can give zero net force while each face experiences nonzero pressure.

Kavg,A = Kavg,B = (3/2)kB(300) = 6.21 × 10⁻²¹ J.

vB,rms/vA,rms = √(mA/mB) = 1/2. Both average vector velocities can be zero.

The hotter peak moves to larger speed, its height decreases, its width and high-speed tail increase, and its total area stays one.

v_rms,new/v_rms,old = √(1200/300) = 2.

n = P₁V₁/(RT₁) = 600/(8.31 × 300) = 0.241 mol.

n is fixed because the sample is sealed. P₂ = P₁(T₂/T₁)(V₁/V₂) = (1.50 × 10⁵)(1.5)(0.5) = 1.125 × 10⁵ Pa.

Random instantaneous velocities, negligible particle volume, elastic collisions, and appreciable forces only during collisions define this classical model.

At high density particle volumes may matter; near condensation intermolecular attractions can persist between collisions.

Plot the four points with axes P in kPa and T in °C. Slope = 11/30 = 0.3667 kPa/°C, so P ≈ 100 + 0.3667T_C.

Set P = 0: T_C = −100/0.3667 ≈ −273 °C.

The measured range is 0–90 °C; the intercept is far outside it, and a real gas may condense first.

At fixed n and T, P = nRT/V, a decreasing hyperbola on P versus V.

Plot P against 1/V. The line goes through the ideal origin and has slope nRT. Knowing T lets the slope determine n.

Vacuum-separated plates can exchange radiation, so thermal contact is possible without physical touching.

For water Q = −120 J, cooling; for spoon Q = +120 J, heating.

Net thermal transfer is from higher to lower temperature. Total mass or total internal energy alone does not decide this direction.

Atoms in the 400 K sample have larger average kinetic energy. Higher-energy atoms are statistically more likely to give energy to lower-energy atoms, producing net transfer toward the 250 K sample.

One interaction may transfer energy in the opposite direction; the law concerns the spontaneous net transfer over many interactions.

Repeated exchange leads to a common temperature between 250 K and 400 K, with no net thermal transfer. Microscopic exchanges persist.

The combined energy remains constant because the outer boundary is insulated and no work occurs. The exact common T depends on the heat capacities; 325 K is guaranteed only for equal heat capacities.

U_i = (3/2)(0.20)(8.31)(300) = 748 J. Q = +600 J; W_on = −200 J; ΔU = 400 J.

ΔT = 2(400)/[3(0.20)(8.31)] = 160 K, giving T_f ≈ 460 K.

Random atomic kinetic energy increases without an increase in center-of-mass motion.

Ideal gas atoms have no appreciable conservative forces between collisions, and their internal structure is omitted. Only translation contributes in this monatomic model.

A solid’s interactions and arrangement can contribute potential energy, so its internal energy can change through atomic motion and structure even when its center of mass is stationary.

For an isolated combined system, total energy remains constant despite energy redistributions among its parts.

Path H has a vertical segment followed by a horizontal expansion at 300 kPa; path L has the low-pressure expansion followed by a vertical segment.

W_H = −(300 kPa)(2.0 L) = −600 J; W_L = −(100 kPa)(2.0 L) = −200 J. Vertical work is zero.

U = (3/2)PV, so ΔU = (3/2)[(300)(3.0) − (100)(1.0)] = 1200 J.

Q = ΔU − W_on gives Q_H = 1800 J and Q_L = 1400 J. Vertical segments are isovolumetric and horizontal segments are isobaric.

The isotherm through A satisfies PV = 100 kPa·L, or P(kPa) = 100/V(L), a decreasing hyperbola. Neither full path maintains that product.

For a quasistatic expansion doing nonzero work against external pressure, isothermal ideal-gas expansion has ΔU = 0 and Q = −W_on > 0. Adiabatic expansion has Q = 0, ΔU = W_on < 0, and decreasing T. An insulated free expansion into vacuum has W_on = 0 instead, so ΔU = 0 and the final ideal-gas T is unchanged.

ΔT_A = 3600/(0.20 × 900) = 20 K; ΔT_B = 3600/(0.20 × 450) = 40 K.

Both lines go through the origin. Slopes are m c_A = 180 J/K and m c_B = 90 J/K.

The larger sample still has c_A = 900 J/(kg·K), but mc = 540 J/K, three times the original sample heat capacity.

Atomic arrangement and interactions govern energy storage. For this course, treat specific heat as temperature independent.

Rate = kAΔT/L = (0.40)(2.0)(15)/(0.10) = 120 W from the 35 °C face toward the 20 °C face.

|Q| = (120)(300) = 3.6 × 10⁴ J. Replacement factor = 2(1/2)/3 = 1/3; rate = 40 W.

Use equal A and L, maintain the same measured face ΔT, insulate side faces, wait for steady state, and measure energy transferred per time with a calibrated sensor or thermal power needed to maintain a boundary.

k = rate × L/(AΔT) is intrinsic under the model and reflects atomic arrangement and interactions. A greater L reduces rate without changing k.

c concerns energy per mass per temperature change; k concerns rate through a temperature gradient. Their units and physical meanings differ.

Q = 0 because the boundary is insulated; W_on = 0 because opposing external pressure is zero; ΔU = 0 by the first law. Fixed n then gives unchanged final T.

Entropy increases because the gas and its energy are more dispersed across the available volume and more microscopic arrangements represent that macrostate.

At fixed isolated constraints, equilibrium has the greatest multiplicity of accessible microscopic arrangements and no spontaneous macroscopic change.

Macroscopic return would concentrate the gas in fewer accessible arrangements and decrease isolated entropy. It is not a spontaneous thermodynamic process.

The block is closed: no matter leaves, but energy leaves thermally. Its entropy can decrease.

The room receives energy and gains entropy. For an isolated boundary including block and surroundings, total entropy cannot decrease and increases for finite-temperature-difference direct heat transfer.

The block’s entropy change is identical for both because its initial and final states are identical.

The ideal reversible combined process has zero total entropy change; the ordinary irreversible route produces positive total entropy change. Reversible means undoable without leaving a net change in system and surroundings.

PV values at A,D,C,B are 100,300,900,300 J. Since n is fixed, temperatures rank T_C > T_D = T_B > T_A.

AD and CB have no work. DC gives W_on = −600 J; BA gives +200 J. Net W_on = −400 J.

The cycle returns to the initial state, so net ΔU = 0 and net Q = −W_on = +400 J.

U_A = 150 J, U_D = 450 J, U_C = 1350 J, U_B = 450 J. Segment ΔU values are +300,+900,−900,−300 J.

Using Q = ΔU − W_on gives Q_AD = +300 J, Q_DC = +1500 J, Q_CB = −900 J, Q_BA = −500 J.

Total heating is 1800 J; total cooling magnitude is 1400 J; their difference is 400 J.

The cyclic engine returns to its original state, so its entropy change is zero. The single reservoir loses thermal energy and has an entropy decrease. The ideal lifted-weight work store gains organized gravitational potential energy without a compensating entropy increase in this idealization.

For the isolated engine–reservoir–ideal-work-store total, the engine has zero entropy change and the ideal work store contributes no compensating entropy increase, leaving the reservoir’s decrease as a negative total. That contradicts the second law.

Weigh the dry metal and water separately; equilibrate metal in the measured warm bath; record initial water and metal temperatures; transfer quickly into a lidded insulated cup, gently stir, and log temperature until a common stable maximum is established. Use a separate measured water mass each trial.

Choose metal plus water and cup. If cup and leakage are negligible, m_m c_m(T_f − T_m) + m_w c_w(T_f − T_w) = 0.

c_m = (0.200)(4200)(23.0 − 20.0)/[(0.100)(80.0 − 23.0)] = 2520/5.70 ≈ 442 J/(kg·K).

At T_f = 22.8 °C, c = (8400)(2.8)/(57.2) ≈ 411 J/(kg·K). At 23.2 °C, c = (8400)(3.2)/(56.8) ≈ 473 J/(kg·K).

The water rises only 3.0 °C, so ±0.2 °C is a substantial fraction of that small difference. The metal’s 57 °C fall is much less sensitive fractionally.

Neglecting cup absorption credits the metal with only the water’s gain even though the metal supplied more energy; this underestimates c.

Total energy remains constant while energy moves from higher-temperature metal to lower-temperature water and cup. Combined entropy increases during this finite-temperature-difference irreversible transfer.

At the common equilibrium temperature, microscopic exchanges continue but there is no net thermal transfer; entropy is maximal under the isolated constraints.

Initial rate = kAΔT/L = (0.10)(0.020)(100)/(0.010) = 20 W toward the cooler gas.

As the gas warms, the face-to-face difference falls, so the rate need not remain 20 W. Measure the rate over time or the net transferred energy, or justify a measured average rate.

At fixed P, PΔV = nRΔT. Thus W_on = −nRΔT and ΔU = (3/2)nRΔT. The first law gives Q = (5/2)nRΔT.

ΔT = 2Q/(5nR) = 360/[5(0.50)(8.31)] = 17.3 K. W_on = −nRΔT = −72 J and ΔU = 108 J.

v_rms,f/v_rms,i = √(317.3/300) ≈ 1.028.

The warming and expanding gas gains entropy, while the hot bath loses entropy as it supplies energy.

Heat transfer across a finite temperature difference is irreversible, so the isolated combined entropy increases even with ideal mechanical work transfer to the piston load.